If $\int \frac{dx}{1 + \sin x} = \tan\left(\frac{x}{2} + a\right) + b$, then the value of $-\frac{4a}{\pi}$ must be
Step-by-Step Solution
Key Concept: Recognize that $1 + \sin x$ is a perfect square when expressed in terms of $\sin\frac{x}{2}$ and $\cos\frac{x}{2}$, simplifying the integrand significantly.
Rewrite $1 + \sin x = 1 + 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2$. Let $t = \tan\frac{x}{2}$, so $2dt = \sec^2\frac{x}{2}dx$. The integral becomes $I = -\frac{2}{(1+t)} + c = -\frac{2}{1+\tan\frac{x}{2}} + c = \frac{\tan\frac{x}{2} - 1}{\tan\frac{x}{2} + 1} + c - 1$.
Correct Answer: 2,4